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3 years ago

What is the value of this expression when x=-1 and y=2 4x3y3

Mathematics

1 answer:

ddd [48]3 years ago

5 0

I believe it is:

4 x -1 x 3 x 2 x 3
4 x 3 x 2 x 3
12 x 6
72

Hope this helped

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nasty-shy [4]

Answer:

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Step-by-step explanation:

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7 0

3 years ago

Use (r,16) to solve y=2x-4

damaskus [11]

(x,y) = (r,16)

With this knowledge, plug in r for x, and 16 for y

16 = 2(r) - 4

Isolate the r, add 4 to both sides

16 (+4) = 2r - 4 (+4)

20 = 2r

Divide 2 from both sides

20/2 = 2r/2

r = 20/2

r = 10

hope this helps

4 0

3 years ago

Simplify the difference (-7×-5×^4+5)-(-7×^4-5-9×)

padilas [110]

-7x-5v^4+5+7x^4+5+9x-=
2x^4+2x+10

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4 years ago

Find the domain and range of the exponential function h(x) = 125 x

NARA [144]

Domain is all real numbers.
Range is all real numbers geater than 0

8 0

4 years ago

1) Use power series to find the series solution to the differential equation y'+2y = 0 PLEASE SHOW ALL YOUR WORK, OR RISK LOSING

iogann1982 [59]

$y=\displaystyle\sum_{n=0}^\infty a_nx^n$

then

$y'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n$

The ODE in terms of these series is

$\displaystyle\sum_{n=0}^\infty(n+1)a_{n+1}x^n+2\sum_{n=0}^\infty a_nx^n=0$

$\displaystyle\sum_{n=0}^\infty\bigg(a_{n+1}+2a_n\bigg)x^n=0$

$\implies\begin{cases}a_0=y(0)\\(n+1)a_{n+1}=-2a_n&\text{for }n\ge0\end{cases}$

We can solve the recurrence exactly by substitution:

$a_{n+1}=-\dfrac2{n+1}a_n=\dfrac{2^2}{(n+1)n}a_{n-1}=-\dfrac{2^3}{(n+1)n(n-1)}a_{n-2}=\cdots=\dfrac{(-2)^{n+1}}{(n+1)!}a_0$

$\implies a_n=\dfrac{(-2)^n}{n!}a_0$

So the ODE has solution

$y(x)=\displaystyle a_0\sum_{n=0}^\infty\frac{(-2x)^n}{n!}$

which you may recognize as the power series of the exponential function. Then

$\boxed{y(x)=a_0e^{-2x}}$

7 0

3 years ago